太和一中2017高考成绩:一道初一数学趣味题

原题：设2002x^3=2003y^3=2004z^3，x>0,y>0,z>0,且
3√2002x^2+2003y^2+2004z^2=3√2002+3√2003+3√2004
求1/x+1/y+1/z
（希望是加号，不然白算了！）

2002x^3=2003y^3=2004z^3=k
2002x^2=k/x
2003y^2=k/y
2004z^2=k/z
3√(2002x^2+2003y^2+2004z^2)=3√(k/x+k/y+k/z)=3√[k(1/x+1/y+1/z)]

2002x^3=2003y^3=2004z^3=k
3√k=(3√2002)x=(3√2003)y=(3√2004)z
(3√2002)=(3√k)/x,(3√2003)=(3√k)/y,(3√2004)=(3√k)/z
3√2002+3√2003+3√2004=(3√k)(1/x+1/y+1/z)

3√[k(1/x+1/y+1/z)]=(3√k)(1/x+1/y+1/z)
k(1/x+1/y+1/z)=k(1/x+1/y+1/z)^3
(1/x+1/y+1/z)^2=1
1/x+1/y+1/z=1

2002x^3=2003y^3=2004z^3=k
2002x^2=k/x
2003y^2=k/y
2004z^2=k/z
3√(2002x^2+2003y^2+2004z^2)=3√(k/x+k/y+k/z)=3√[k(1/x+1/y+1/z)]

2002x^3=2003y^3=2004z^3=k
3√k=(3√2002)x=(3√2003)y=(3√2004)z
(3√2002)=(3√k)/x,(3√2003)=(3√k)/y,(3√2004)=(3√k)/z
3√2002+3√2003+3√2004=(3√k)(1/x+1/y+1/z)

3√[k(1/x+1/y+1/z)]=(3√k)(1/x+1/y+1/z)
k(1/x+1/y+1/z)=k(1/x+1/y+1/z)^3
(1/x+1/y+1/z)^2=1
1/x+1/y+1/z=1